Question

If p/q=(3/2)-squre ÷(6/7),then find the value of (p/q)-cube​

Answer 1

$\frac{p}{q} = {( \frac{3}{2} )}^{ - 2} \div \frac{6}{7} \\ \frac{4}{9} \times \frac{7}{6} \\ \frac{28}{54} \\ \frac{14}{27}$
$\frac{p}{q} = \frac{14}{27} \\ {( \frac{p}{q} )}^{ - 3} = {( \frac{14}{27} )}^{ - 3} \\ {( \frac{27}{14} )}^{3}$

Answer 2

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